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Standard IX

Physics

Longitudinal Waves

The number of...

Question

The number of particles are given by- n= -D(n2-n1)/(x2-x1) are crossing the unit area perpendicular to x-axis in unit time,n1 and n2 are the number of particles per unit volume for the values of x meant to be x1 and x2. how to calculate the dimensional formula of diffusion constant D?

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Solution

Given, n = $- D (\frac{n_{2} - n_{1}}{x_{2} - x_{1}})$
$D = \frac{n (x_{2} - x_{1})}{- (n_{2} - n_{1})} D i m e n s i o n s o f n = n u m b e r / a r e a / t i m e = L^{- 2} T^{- 1} D i m e n s i o n s o f n_{1}, n_{2} = n u m b e r / v o l u m e = L^{- 3} D i m e n s i o n o f x_{2} - x_{1} = d i f f . b e t w e e n t w o p o s i t i o n s = L T h e r e f o r e, D = \frac{(L^{- 2} T^{- 1}) (L)}{(L^{- 3})} = [L^{2} T^{- 1}]$

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The number of particles are given by- n= -D(n2-n1)/(x2-x1) are crossing the unit area perpendicular to x-axis in unit time,n1 and n2 are the number of particles per unit volume for the values of x meant to be x1 and x2. how to calculate the dimensional formula of diffusion constant D?

Given, n = - Dn2 - n1x2 - x1 D = n (x2 - x1)- (n2 - n1)Dimensions of n = number/area/time = L-2 T-1Dimensions of n1, n2 = number/volume = L-3Dimension of x2 - x1 = diff. between two positions = LTherefore, D = (L-2T-1) (L)( L-3) = [L2T-1]